32 MATHEMATICS
+ ✁✁ + ✂✂ =
- b
a
,
✁ + ✁✁✂ + ✂ =
c
a, ✁✁✂ =- d
a
.
Let us consider an example.
Example 5* : Verify that 3, –1,
1
3
✄ are the zeroes of the cubic polynomialp(x) = 3x^3 – 5x^2 – 11x – 3, and then verify the relationship between the zeroes and the
coefficients.
Solution : Comparing the given polynomial with ax^3 + bx^2 + cx + d, we get
a = 3, b = – 5, c = –11, d = – 3. Further
p(3) = 3 × 3^3 –^ (5 × 3^2 ) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)^3 – 5 × (–1)^2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,11 1 1^32
35113
33 3 3p☎✠✝ ✡✆✞ ✟✠☎✝ ✡✆ ✝ ✟✠☎✝ ✡✆ ✝ ✟✠☎✝ ✆✡✝
☛ ☞ ☛ ☞ ☛ ☞ ☛ ☞,
=
–^1511 3–^220
99 3 33
✄ ✌ ✄ ✍ ✌ ✍
Therefore, 3, –1 and
1
3
✄ are the zeroes of 3x^3 – 5x^2 – 11x – 3.So, we take = 3, ✁ = –1 and ✂ =
1
3
✄ ✎
Now,
11 5(5)
3(1) 2
3333
b
a✑ ✒ ✏✏ ✏
✓✔✕✔✖✗ ✔ ✏ ✔✘✏ ✙✗ ✏ ✗ ✗ ✗
✚ ✛
,
11 1 1 1
3(1) (1) 3 3 1
33 3 3
c
a✑ ✒ ✑ ✒ ✏
✓✕✔✕✖✔✖✓✗ ✜ ✏ ✔ ✏ ✜✘✏ ✙✔✘✏ ✙✜ ✗✏ ✔ ✏ ✗ ✗
✚ ✛ ✚ ✛
,
3(1) 1( 1 3)
33
d
a✥✦✧★ ✩ ✢ ✩✣✢ ✤★ ★✢✢ ★✢
✪ ✫
✬ ✭
.
*Not from the examination point of view.